$h \in C^k$ implies $\frac{h(x)-h(0)}{x} \in C^{k-1}$

Question

This is a self-answered question. I post it here since it wasn't obvious to me. (although I have seen similar questions-is it an exact duplicate?).

Let $h:\mathbb{R}\to\mathbb{R}$ be $C^k$ and suppose that $h(0)=0$. Define $$ F(x) = \begin{cases} \frac{h(x)}{x} & \text{if $x\neq 0$} \\ h'(0) & \text{if $x=0$}\end{cases} $$

Then $F$ is $C^{k-1}$ and $F^{(m)}(0)=\frac{h^{(m+1)}(0)}{m+1}$ for every $m \le k-1$.

Is there a way to deduce the equality $F^{(m)}(0)=\frac{h^{(m+1)}(0)}{m+1}$ without a long calculation via L'Hôpital's rule? Perhaps using approximation by Taylor's polynomials?

I tried to that, but naive attempt didn't make it. (By inspecting the Taylor's polynomials, it is immediate that if $F \in C^{k-1}$, then $F^{(m)}(0)=\frac{h^{(m+1)}(0)}{m+1}$ holds, but this does not imply the $C^{k-1}$ differentiability to begin with).

Added:

This result is sharp. Indeed $h(x)=x^{k+1}\text{sgn}(x)$ is $C^k$, but $F(x)=x^{k}\text{sgn}(x)$ is only $C^{k-1}$, and is not differentiable $k$ times at zero.

Answer 1

The proof is essentially taken from this answer.

We shall prove the claim, by finite induction-i.e. we shall prove that $F$ is $C^m$ for $m=0,1,\dots,k-1$.

We shall use L'Hôpital's rule to find $\lim_{x\to 0} F^{(m)}(x)$.

By the general Leibniz rule, $$ \begin{align*} F^{(m)}(x) &= \sum_{k=0}^m \binom{m}{k} h^{(m-k)}(x)\cdot(-1)^kk!x^{-k-1} \\ &= \frac{\sum_{k=0}^m (-1)^k k!\binom{m}{k}x^{m-k}h^{(m-k)}(x)}{x^{m+1}}. \end{align*} $$

All the summands in the numerator, tend to zero when $x \to 0$; $\lim_{x \to 0}x^{m-k}=0$ for $k <m$, and for $k=m$, we have $\lim_{x \to 0}h^{(m-k)}(x)=\lim_{x \to 0}h(x)=h(0)=0$.

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Thus, the limit is of the indeterminant form $\frac{0}{0}$, and by L'Hôpital's rule we get $$ \begin{align} \lim_{x\to 0} F^{(m)}(x) &= \lim_{x\to 0} \frac{\sum_{k=0}^{m-1} (-1)^kk!\binom{m}{k}(m-k)x^{m-k-1}h^{(m-k)}(x) + \sum_{k=-1}^{m-1}(-1)^{k+1}(k+1)!\binom{m}{k+1}x^{m-k-1}h^{(m-k)}(x)}{(m+1)x^m} \\ &= \lim_{x\to 0} \frac{x^mh^{(m+1)}(x)}{(m+1)x^m} = \frac{h^{(m+1)}(0)}{m+1}. \end{align}, $$ where in the passage to the third line we have used the fact that all terms in both sums cancel each other, except the $k=-1$ term in the second summand.

If we assume (by induction) that $F^{(m-1)}(x)$ is continuous, then this claim shows that $F^{(m)}(0)$ exists and equal to $\lim_{x\to 0} F^{(m)}(x)$. Thus $F^{(m)}(x)$ is continuous at $x=0$.

This finishes the proof.

Answer 2

We have $$F(x) = \int_{0}^1 h'(tx) dt \,.$$ When $h$ is $C^k$, all partial derivatives w.r.t. $x$ of $(t, x) \mapsto h'(tx)$ up to order $k-1$ are continuous. Hence by the dominated convergence theorem and the mean value theorem, we may change the order of integration and differentiation, and conclude that $F$ is $C^{k-1}$.